Question 1 |

In an equilateral triangle PQR, side PQ is divided into four equal parts, side QR is divided into six equal parts and side PR is divided into eight equals parts. The length of each subdivided part in cm is an integer. The minimum area of the triangle PQR possible, in cm^2, is

18 | |

24 | |

48 \sqrt{3}
| |

144 \sqrt{3} |

Question 1 Explanation:

For \left(\frac{a}{4}, \frac{a}{6}, \frac{a}{8}\right) to be integer, a must be LCM of 4, 6 and 8. So a = 24

\text { Area }=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4} \times 24^{2}=144 \sqrt{3}

Question 2 |

In the figure shown above, PQRS is a square. The shaded portion is formed by the intersection of sectors of circles with radius equal to the side of the square and centers at S and Q.

The probability that any point picked randomly within the square falls in the shaded area is __________

4-\frac{\pi}{2} | |

\frac{1}{2} | |

\frac{\pi}{2}-1 | |

\frac{\pi}{4} |

Question 2 Explanation:

\begin{aligned} \text { Probability } &=\frac{f A}{T A} \\ f A &=\left(\frac{\pi r^{2}}{4}-\frac{r^{2}}{2}\right) \times 2 \\ \frac{f A}{T A} &=\frac{\left(\frac{\pi r^{2}}{4}-\frac{r^{2}}{2}\right) \times 2}{r^{2}}=\left(\frac{\pi}{2}-1\right) \end{aligned}

Question 3 |

On a planar field, you travelled 3 units East from a point O. Next you travelled 4 units South to arrive at point P. Then you travelled from P in the North-East direction such that you arrive at a point that is 6 units East of point O. Next, you travelled in the North-West direction, so that you arrive at point Q that is 8 units North of point P.

The distance of point Q to point O, in the same units, should be _____________

The distance of point Q to point O, in the same units, should be _____________

3 | |

4 | |

5 | |

6 |

Question 3 Explanation:

O Q=\sqrt{3^{2}+4^{2}}=5

Question 4 |

Four persons P, Q, R and S are to be seated in a row. R should not be seated at the second position from the left end of the row. The number of distinct seating arrangements possible is:

6 | |

9 | |

18 | |

24 |

Question 4 Explanation:

Number of arrangements =3 \times 3 !=18

Question 5 |

\oplus and \odot are two operators on numbers p and q such that p \odot q=p-q, and p \oplus q=p \times q

Then, (9 \odot(6 \oplus 7)) \odot(7 \oplus(6 \odot 5))=

Then, (9 \odot(6 \oplus 7)) \odot(7 \oplus(6 \odot 5))=

40 | |

-26 | |

-33 | |

-40 |

Question 5 Explanation:

\begin{aligned} [9-(6 \times 7)]-[7 \times 1] &=-33-7 \\ &=-40 \end{aligned}

Question 6 |

Two identical cube shaped dice each with faces numbered 1 to 6 are rolled simultaneously. The probability that an even number is rolled out on each dice is:

\frac{1}{36} | |

\frac{1}{12} | |

\frac{1}{8} | |

\frac{1}{4} |

Question 6 Explanation:

Probability of getting even number on a dice=\frac{3}{6}=\frac{1}{2}

\thereforeTwo dice are rolled simultaneously,

Hence required probability =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}

\thereforeTwo dice are rolled simultaneously,

Hence required probability =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}

Question 7 |

The mirror image of the above text about X-axis is

A | |

B | |

C | |

D |

Question 8 |

A function, \lambda, is defined by

\lambda(p, q)=\left\{\begin{array}{cl} (p-q)^{2}, & \text { if } p \geq q \\ p+q, & \text { if } p \lt q \end{array}\right.

The value of the expression \frac{\lambda(-(-3+2),(-2+3))}{(-(-2+1))} is:

\lambda(p, q)=\left\{\begin{array}{cl} (p-q)^{2}, & \text { if } p \geq q \\ p+q, & \text { if } p \lt q \end{array}\right.

The value of the expression \frac{\lambda(-(-3+2),(-2+3))}{(-(-2+1))} is:

-1 | |

0 | |

\frac{16}{3} | |

16 |

Question 8 Explanation:

\frac{\lambda(-(-3+2),(-2+3))}{(-(2+1))}=\lambda \frac{(1,1)}{1}=\lambda(1,1)

So, 1st definition will be applicable as p = q.

\text { Hence, } \qquad \lambda(1,1)=(1-1)^{2}=0

So, 1st definition will be applicable as p = q.

\text { Hence, } \qquad \lambda(1,1)=(1-1)^{2}=0

Question 9 |

Five line segments of equal lengths, PR, PS, QS, QT and RT are used to form a star as shown in the figure above.

The value of \theta, in degrees, is ________

36 | |

45 | |

72 | |

108 |

Question 9 Explanation:

Sum of angle formed at the pentagon = 540^{\circ}

Each angle of =\frac{540}{5}=108^{\circ}

\angle x=180-108=72^{\circ}

Sum of angle of triangle =180^{\circ}

\begin{aligned} 72^{\circ}+72^{\circ}+\theta &=180^{\circ} \\ \theta &=36^{\circ} \end{aligned}

Question 10 |

Consider two rectangular sheets, Sheet M and Sheet N of dimensions 6cm x 4cm each.

Folding operation 1: The sheet is folded into half by joining the short edges of the current shape.

Folding operation 2: The sheet is folded into half by joining the long edges of the current shape.

Folding operation 1 is carried out on Sheet M three times.

Folding operation 2 is carried out on Sheet N three times.

The ratio of perimeters of the final folded shape of Sheet N to the final folded shape of Sheet M is ____.

Folding operation 1: The sheet is folded into half by joining the short edges of the current shape.

Folding operation 2: The sheet is folded into half by joining the long edges of the current shape.

Folding operation 1 is carried out on Sheet M three times.

Folding operation 2 is carried out on Sheet N three times.

The ratio of perimeters of the final folded shape of Sheet N to the final folded shape of Sheet M is ____.

0.546527778 | |

0.126388889 | |

0.295138889 | |

0.217361111 |

Question 10 Explanation:

(\text { Perimeter })_{M}=2(2+1.5)=7

\text { (Perimeter })_{N}=2(0.5+6)=13

Required ratio =\frac{13}{7}

There are 10 questions to complete.

And solution??